[LeetCode] Rising Temperature 上升溫度

 

Given a Weather table, write a SQL query to find all dates' Ids with higher temperature compared to its previous (yesterday's) dates.

+---------+------------+------------------+
| Id(INT) | Date(DATE) | Temperature(INT) |
+---------+------------+------------------+
|       1 | 2015-01-01 |               10 |
|       2 | 2015-01-02 |               25 |
|       3 | 2015-01-03 |               20 |
|       4 | 2015-01-04 |               30 |
+---------+------------+------------------+

For example, return the following Ids for the above Weather table:

+----+
| Id |
+----+
|  2 |
|  4 |
+----+

 

這道題給了我們一個Weather表，讓我們找出比前一天溫度高的Id，由於Id的排列未必是按順序的，所以我們要找前一天就得根據日期來找，我們可以使用MySQL的函數Datadiff來計算兩個日期的差值，我們的限制條件是溫度高且日期差1，參見代碼如下： 

 

解法一：

SELECT w1.Id FROM Weather w1, Weather w2
WHERE w1.Temperature > w2.Temperature AND DATEDIFF(w1.Date, w2.Date) = 1;

 

下面這種解法我們使用了MySQL的TO_DAYS函數，用來將日期換算成天數，其余跟上面相同：

 

解法二：

SELECT w1.Id FROM Weather w1, Weather w2
WHERE w1.Temperature > w2.Temperature AND TO_DAYS(w1.Date) = TO_DAYS(w2.Date) + 1;

 

我們也可以使用Subdate函數，來實現日期減1，參見代碼如下：

 

解法三：

SELECT w1.Id FROM Weather w1, Weather w2
WHERE w1.Temperature > w2.Temperature AND SUBDATE(w1.Date, 1) = w2.Date;

 

最后來一種完全不一樣的解法，使用了兩個變量pre_t和pre_d分別表示上一個溫度和上一個日期，然后當前溫度要大於上一溫度，且日期差為1，滿足上述兩條件的話選出來為Id，否則為NULL，然后更新pre_t和pre_d為當前的值，最后選出的Id不為空即可：

 

解法四：

SELECT Id FROM (
SELECT CASE WHEN Temperature > @pre_t AND DATEDIFF(Date, @pre_d) = 1 THEN Id ELSE NULL END AS Id,
@pre_t := Temperature, @pre_d := Date 
FROM Weather, (SELECT @pre_t := NULL, @pre_d := NULL) AS init ORDER BY Date ASC
) id WHERE Id IS NOT NULL;

 

參考資料：

https://leetcode.com/discuss/33641/two-solutions

https://leetcode.com/discuss/52370/my-simple-solution-using-inner-join

https://leetcode.com/discuss/86435/a-simple-straightforward-solution-and-its-very-fast

 

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